Download (1) and (2)

Chapter 5
Static Output Feedback and
Estimators
Because state feedback uses all the information of
the system, it is also called full information feedback. We
can often get a perfect result in theory by state feedback .
Output feedback uses an incomplete information of
the system. Hence, it is much more difficult to stabilize a
system or improve the performance by using static output
feedback.
v
B
x

A
K
x
C
y
In this chapter, we first introduce the knowledge of
pole-placement using static output feedback. Then, we will
study the design of estimators.
§5-1 Static Output Feedback and Pole Placement
1. Property of static output feedback
Consider linear time-invariant system as follows:
x  Ax  Bu
(5-1)
y  Cx
Let
u=Ky+v
(5-2)
In general, Eq. (5-2) is said to be a static output
feedback control law, where K is a p×q constant
matrix, v is a p-dimensional vector.
From Eq. (5-1) and Eq. (5-2), we easily obtain the
dynamical equation of the closed-loop system as
follows:
x  (A  BKC)x  Bv , y  Cx
x
x
C
v
B

(5  3)
y
A
K
Closed-loop system
Theorem 5-1 Feedback control law (5-2) does not change
the observability of the system.
Proof Firstly, we have the following equation
 sI  ( A  BCK )   I BK   sI  A 



 C 
C
0
I

 


(5-4)
Because the right hand side of Eq. (5-4) is
nonsingular, for any s and K, we have
s I  ( A  BKC) 
s I  A 
rank 
 rank 


C
C




(5  5)
Q.E.D
Hence, system (A+BKC, C) is observable iff (A, C) is
observable, which implies that static output feedback
does not change the obervability of the system.
Suppose system (A, C) is unobservable. From Eq. (5-5),
the s that will reduce rank of the right hand side of Eq.(55) also reduces the rank of the left hand side. Thus, we
conclude that static output feedback does not change
Corollary: Feedback control law u=Ky+v does not
change the controllability of systems.
In chapter 4, it is proved that poles of a controllable
system can be arbitrarily assigned. Unfortunately, output
feedback as a special case of state feedback does not
have such a property in general.


x1
1
s
x2
1
s
Poles can be
arbitrarily
assigned by
state feedback.
k1
k2

1
s
x1
1
s
x2
k2
The relation between state
Poles cannot
be arbitrarily
assigned by
output
feedback.
2. Cyclic Matrix
1. Definition of cyclic matrix
Definition: n×n square matrix A is called a cyclic matrix
if its minimal polynomial is its character polynomial. The
following statements are equivalent:
1). There only exists a non-one invariant factor for Smith
canonical form of sIA;
2). An eigenvalue only has one Jord block in Jordancanonical-form of A.
3 1 0

0 3 1

Example: Consider matrix



A  0 0 3


2 1

It is clear that A is a cyclic

0 2 
matrix.
In particular, we have
1) Suppose the eigenvalues of A are distinct. Then A is a
cyclic matrix if there is only one Jordan block for
each eigenvalue.
2) A is a cyclic matrix iff there exists a vector b such
that
b, Ab, A2 b,
, An 2 b, An 1b
are linearly independent, i.e. (A, b) is controllable.
Hence, for single input system (A, b), it is
controllable iff A is a cyclic matrix and b is a generated
element of A.
2. Cyclic matrix and controllability
Corollary 1*: Suppose (A, B) is controllable and A is a
cyclic matrix. Then for almost all real number and r
)
suitable vector, ( A, Br )  ( Ais, b
controllable.
Remark: The above corollary means that if A is a cyclic
matrix, the design procedure to obtain
1
K1  u1 u 2
un 1 0x1 x 2
x n n n  R pn
can be omitted.
For example, the following system is controllable:
3 1 0

0 3 1




A  0 0 3


2 1


0 2 
0
0

B  2

4
 2
 
4
 

0
 1   
1  , b:=B  B      

 2   
3
 
 
0 
Generally, consider
 J1

J2

A




 B1 

B 
,B   2 

 

 
Js 
 Bs 
(A, B) is controllable
That A is a cyclic
matrix means if i  j
then l i  l j
.
Bi
The last row of hisi nonzero.

From the above analysis, we have for
p 1
hi r  0, i  1, 2, , s .
r

R
almost any
,
（A, B )  ( Ais
, bcontrollable
）
The
last row
nonzero.
of ish i
Bi
4) Lemma 2*: Suppose (A, B) is controllable. Then for
K  R pn of A+BK are distinct.
almost all
, eigenvalues
Hence, A+BK is a cycle matrix.
Refer to (Davison, 1968, IEEE AC, No.6) for the proof.
3. Theorems used in output feedback poleplacement
Corollary 5-2 Suppose (A, B) is controllable and A is a
cyclic matrix. Then there exists a vector bImB such
that (A, b) is controllable.
Corollary 5-4 Suppose (A, B, C) is controllable and
observable. Then there exists a p×q matrix H such
that (A+BHC, B) is controllable and (A+BHC, C) is
observable. Moreover, A+BHC is a cyclic matrix, i.e.
the order of its minimum polynomial is n.
Example 5-3 Consider system (A, B, C):
1
0
A
0

0
1
1
0
0
0
1
1
0
0
0 
0

1
0 1 
H

0
0


0
0
B
1

0
0
0 
0

1
1
0
A  BHC  
0

0
b   0 0 0 1
T
1 0 0 0 
C

0
0
0
1


1
1
0
0
0
1
1
0
0
0 
1

1
c  1 0 0 0
It is easy to verify that (A+BHC, B) is controllable,
(A+BHC, C) is observable and A+BHC is a cyclic
matrix. Moreover, the order of its minimum polynomial
is 4.
4. Pole placement with static output
feedback
1. Conditions that n poles can be assigned for single
input systems
First of all, let us study a SIMO system to get an idea
of the difficulties of pole placement with static output
feedback, which do not arise in state feedback systems.
x  Ax  bu,
A  R nn , b  R n1
y  Cx, C  R qn
u  Kx  v , K  [k1 k 2
（5-11）
kq ]  R1q
(5  12)
The dynamical equation of closed-loop system is
x  (A  bKC)x  bv
(5  13)
Let the characteristic equation of A and A+bKC be △o(s)
and △c(s), respectively.
o (s )  s n  an 1s n 1  an 2s n 2 
 a1s  a 0
c (s )  s n  an 1s n 1  an 2s n 2 
 a1s  a 0
！
If (A, b) is controllable, then it can be transformed into a
canonical form with a matrix P
 0


1
A  PAP  


 a0
1
1
a1





1 
an 1 
C  CP 1
0
 
b  Pb   
0
 
1 
Now the closed-loop system matrix becomes
1
 0




 0
 a 0 a1
1
1
 0
 0

  

  
    KC  

  
1  0
 0
 a 0 a 1
an 1  1 

1





1 
a n 1 
q  1 matrix and
Denote C  [ c1 c2 , cwhere
is a ci column
n]
Kci
a scalar.
ai  ai  Kci 1
(i  0,
n  1)
For a set of given closed-loop poles, isadetermined.
i
What we shall do is to choose a K1q, such that
is
c1T KT  a 0  a 0
c2T KT  a1  a1
(5-14)
cnT KT  an 1  an 1
or
T
C KT  d
（5  15）
cq1   k1   a 0  a 0 
cq 2  k 2   a1  a1 


   

  

  

cqn  kq  an 1  an 1 
c11 c21
c
 12 c22


c2n 1

c1n c2n

CT
KT
d
c11 c21
c
 12 c22


c2n 1

c1n c2n

cq1   k1   a 0  a 0 
cq 2  k 2   a1  a1 


   

  

   aq 1  aq 1 

cqn  kq  
CT
KT
d
If the desired poles satisfy the nq equations, then it is
possible to assign the poles with output feedback.
Example 5-4
0 1 0
0
1 0 2 




A   0 0 1  , b  0  , C  
, K=[k1 k 2 ]

0 1 1

 3 2 4 
1 
 s 3  4s 2  2s  3; where rankC=q=2, from the
Condition 5-15, we have
1 0
 3- a 0  k1  3- a 0 (1)
 k1  




T
T
C K  0 1      2- a1   k 2  2- a1 (2)
k2 

 2 1 
4- a 2  
2k1  k 2  4- a 2
(3)
Substituting (1) and (2) into (3), then we get the
restriction as follows:
2a 0  a1  a 2  4
(5  16)
Hence, if the desired poles satisfy the above equation,
then they can be assigned. From the above equation we
have:
2(l 1l 2l 3 )  (l 1l 2  l 1l 3  l 2l 3 )  (l 1  l 2  l 3 )  4
a0
a1
 a2
It is easy to verify that the set of poles {1, 1, 2}
can be assigned by output feedback, but {5/6, 1, 2}
can not be assigned by output feedback.
ai  ai  Kci 1
(i  0,
n  1)
the characteristic equation of the closed-loop system is
s n  (an 1  K cn )s n 1  (an  2  K cn 1 )s n  2 

(a1  K c2 ) s  (a 0  K c1 )  0
Substituting i(i=1,2,…,q) into the above equation, we
have
l in  an 1l in 1 
 a1l i  a0  K cn l in 1 
 K c2l i  K c1
i
i.e.
o (i )  KChi
i  1, 2,
q
l in  an 1l in 1 
 a1l i  a0  K cn l in 1 
 K[ c1 c2
i.e.
 1 
 l 
i 

cn ]
 KChi


 n 1 
l i 
o (i )  KChi
where
)
Denoting  o (ias
 K c2l i  K c1
i  1,2,
 1 
 l 
i 

hi 


 n 1 
l i 
,
iti follows that
q,
S
[1  2
q ]

 K1q Cqn [h1 h 2
1q
1
 1
 l
l2
 1
 KC 

l 1n 1 l 2n 1
If
h q ]nq  KS qq
S  C[h1 h 2
is nonsingular, then from
=KS
we have
K  [1 2
1 
l q 


n 1
l q 
nq
hq ]
q ]S1
26
Example 5-5
0 1 0
0 
1 0 2 




A   0 0 1  , b  0  , C  

0
1
1


 3 2 4 
1 
Compute output feedback K:
K=[k1 k 2 ]
1）such that the poles are assigned at 1= 1, 2= 2
From the equation:
i =K[ c1 c2
 1 
 l 
i 

cn ]
 KChi , we have


 n 1 
l i 
i  l i3  4l i2  2l i  3  K c3l i2  K c2l i  K c1
then
1
1
h1   1 , h 2   2  , [1  2 ]  [4
 1 
 4 
 1 3 
3 2 
3 9 
1
S  C[h1 h 2 ]  
S 


0 2 
0 1 

2 
1
3
1
 K  [1  2 ]S  [4 7] 
0

3 
2 4
 [
1 3
2 
7]
5
 ]
2
28
2) The desired poles are
l 1  0.5，l 2  0.25
l i3  4l i2  2l i  3  Kc3l i2  Kc2l i  Kc1
1
 
 i =[k1 k 2 ][c1 c2 c3 ]  l i   KChi , S  C[h1 h 2 ]，K  [1  2 ]S 1
l 2 
 i 
then
 1 
 1 
 , h   1/ 4  ,
h1  

1/
2

 2




 1/ 4 

 1/16 

[1  2 ]  [2.875 175 / 64]
1  6 9 / 2 
S= 
 det S=0

4  1 3/ 4 
Let
From the equation:
l 1  0.5  e，l 2  0.25，
l i3  4l i2  2l i  3  Kc3l i2  Kc2l i  Kc1
1
 
 i =[k1 k 2 ][c1 c2 c3 ]  l i   KChi , S  C[h1 h 2 ]，K  [1  2 ]S 1
l 2 
 i 
then


1
 1 


 0.25 ,
h1  
e  0.5
,
h

 2


e 2  e  0.25

 1/16 



[1  2 ]  [e 3  2.5e 2  1.25e  2.875 175 / 64]

1
1 
1 0 2  

S  C[h1 h 2 ]  
e  0.5
0.25


0 1 1   2

1/16
e

e

0.25


K  [1  2 ]S 1
 [e  2.5e  1.25e  2.875
3
2
175 / 64]S
1
1

e
6
e 